Chemistry - Chemical Kinetics Question with Solution | TestHub

$\mathrm{l}\mathrm{o}\mathrm{g}10\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)$
$\mathrm{R}=\mathrm{8.314 }\mathrm{J}{\mathrm{K}}^{-1}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$
${\mathrm{T}}_1=(273+270)\mathrm{K}=\mathrm{543 }\mathrm{K}$
${\mathrm{T}}_2=(273+350)\mathrm{K}=\mathrm{623 }\mathrm{K}$
${\mathrm{k}}_2=4.5\times 10^{-10}{\mathrm{s}}^{-1},{\mathrm{k}}_1=1.3\times 10^{-11}{\mathrm{s}}^{-1}$
${\text{log}}_{10}\frac{4.5\times {10}^{-10}}{1.3\times {10}^{-11}}=\frac{{\text{E}}_{\text{a}}}{2.303\times 8.314}\left(\frac{1}{543}-\frac{1}{623}\right)$
$1.534=\frac{{\mathrm{E}}_{\mathrm{a}}}{19.1471}\left(\frac{80}{543\times 623}\right)$
${\mathrm{E}}_{\mathrm{a}}=\frac{1.534\times 19.1471}{2.364\times 10^{-4}}=\mathrm{124.6 }\mathrm{k}\mathrm{J}$